Sure, just analyze the situation from the point of view of an inertial frame where the accelerating twin's speed during the inertial phase between leaving Earth and turning around is smaller than the Earth's speed in this frame. Naturally if the Earth's speed is greater during this interval, its clock is ticking slower during this interval.
For a numerical example, consider the same situation first viewed from the rest frame of the inertial stay-at-home twin (call this frame A), next viewed from the inertial frame where the traveling twin is at rest between leaving Earth and turning around (call this frame B). If both twins are exactly 30 when the traveling twin departs the stay-at-home twin, and in the stay-at-home twin's frame A, the traveling twin moves away for 10 years at 0.6c before instantaneously accelerating to come to rest relative to the stay-at-home twin, then in frame A the stay-at-home twin turning 40 will happen simultaneously with the traveling twin accelerating while turning 38. But in the frame B of an inertial observer who sees the traveling twin at rest until accelerating, the stay-at-home twin is moving away at 0.6c for 8 years before the traveling twin accelerates (again at age 38) to match speeds with him, and at the moment of acceleration the stay-at-home twin is 36.4 years old. This is just an example of the relativity of simultaneity. Both frames agree the traveling twin is turning 38 at the moment of acceleration, but in the first frame A this event is simultaneous with the stay-at-home twin turning 40, while in the second frame B this event is simultaneous with the stay-at-home twin turning 36.4.
If you doubt these numbers are correct, I can show that they are by using the Lorentz transformation, if you wish.